There are 5 multiples of 18 between 200 and 300. (An even number divisible by 9 is any number divisible by 18).
There are 33 such numbers.
Two of them.
There are infinitely many numbers divisible by 63, just multiply 63 by any whole number and you'll get one.
In order to be divisible by both 2 and 3, a number must be divisible by 6.The largest multiple of 6 less than 100 is 96 ... the 16th multiple of 6.So there are 16 of them.
dont kno so just helpp friendds!
An infinite number.
The first whole number divisible by 3 is 102 and the last one 399. Let n be the number of whole numbers between 102 and 399 102 + (n - 1)x3 = 399 (this is an arithmetic progression) Solving n, n-1 = (399 - 102)/3 = 99 n = 100 Since even whole numbers among these 100 will be divisible by 6, the number not divisible is half of 100, i.e. 50. regards, lpokbeng
There are 5 multiples of 18 between 200 and 300. (An even number divisible by 9 is any number divisible by 18).
There are many numbers between 500 and 1000 divisible by 3 and 9. Any number divisible by 9 is divisible by 3. How about 900?
48 is divisible by ten factors. There is an infinite amount of numbers divisible by 48.
33 numbers between 1 and 100 are divisible by 3.
There are 33 such numbers.
Two of them.
251
18 numbers are there
There are infinitely many numbers divisible by 63, just multiply 63 by any whole number and you'll get one.